HDU_1312——二维空间BFS

这题网上都是用DFS做的,我也不知道怎么用BFS做的。。。有个注意点,就是一步也走不动的时候,输出1

Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 
 

 

Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 
 

 

Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself). 
 

 

Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
 

 

Sample Output
45 59 6 13
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #define MAX 21
 5 using namespace std;
 6 struct point
 7 {
 8    int x,y;  
 9 }start;
10  
11 int N,M,dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
12 char map[MAX][MAX];
13 
14 int Bfs(void);
15 int main()
16 {
17    while(~scanf("%d %d",&M,&N) && N!=0&&M!=0)
18       {
19          memset(map,0,sizeof(map));
20          for(int i=0;i<N;i++)
21             {
22                scanf("%s",map[i]);
23                for(int j=0;j<M;j++)
24                   {
25                      if(map[i][j]=='@')
26                         start.x=i,start.y=j;   
27                   }
28             }
29          int num=Bfs();
30          printf(num==0?"1\n":"%d\n",num);        
31       }
32    return 0;   
33 }
34 
35 int Bfs(void)
36 {
37    int move=0;
38    queue<point>p;
39    p.push(start);
40    point temp,next;
41    while(!p.empty())
42       {
43          temp=p.front();
44          p.pop();
45          for(int i=0,flag=0;i<4;i++)
46             {
47                next.x=temp.x+dir[i][0];
48                next.y=temp.y+dir[i][1];
49                if(next.x>=0&&next.x<N && next.y>=0&&next.y<M && map[next.x][next.y]!='#')
50                   {
51                      map[next.x][next.y]='#';
52                      move++;
53                      p.push(next);
54                   }
55             }
56       }
57    return move;
58 }

 

posted @ 2013-06-11 22:30  瓶哥  Views(174)  Comments(0Edit  收藏  举报